Methyl benzoate is commonly added to perfumes as it has a pleasant smell - Scottish Highers Chemistry - Question 3 - 2018

The small test tube filled with cold water acts as a condenser to prevent the escape of vapors from the reaction mixture. This helps in keeping the reactants contained.

Product X is water.

To determine the limiting reactant, we first calculate the moles of benzoic acid and methanol used.

The molar mass of benzoic acid (C₇H₆O₂) is 122 g/mol, so:

For 5.0 g of benzoic acid:

$ext{moles of benzoic acid} = \frac{5.0 ext{ g}}{122 ext{ g/mol}} = 0.041 ext{ moles}$

The molar mass of methanol (CH₃OH) is 32 g/mol, so:

For 2.5 g of methanol:

$ext{moles of methanol} = \frac{2.5 ext{ g}}{32 ext{ g/mol}} = 0.078 ext{ moles}$

The balanced equation states that one mole of benzoic acid reacts with one mole of methanol. Therefore, 0.041 moles of benzoic acid would need:

$0.041 ext{ moles of } CH₃OH.$

Since we have 0.078 moles of methanol available, the limiting reactant is benzoic acid.

From the experiment, the student produced 3.1 g of methyl benzoate from 5.0 g of benzoic acid. To produce 100 g of methyl benzoate, we can set up a proportion:

$\frac{5.0 ext{ g}}{3.1 ext{ g}} = \frac{x ext{ g}}{100 ext{ g}}$

Thus, solving for x gives:

$x = \frac{5.0 \times 100}{3.1} \approx 161.29 ext{ g of benzoic acid}$

Given that benzoic acid costs £39.80 for 500 g, the cost per gram is:

$\text{Cost per gram} = \frac{39.80}{500} = 0.0796 ext{ £/g}$

Therefore, the cost for 161.29 g is:

$\text{Cost} = 161.29 \times 0.0796 \approx 12.84 ext{ £}$

Rounding this to the nearest penny gives a final cost of approximately £12.84.